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3 years ago

Why is it called endogenous process

Chemistry

1 answer:

anyanavicka [17]3 years ago

7 0

Answer:

The Earth is shaped by many geological processes. ... Processes caused by forces from within the Earth are endogenous processes. Exo is a prefix meaning "out", and endo is a prefix meaning "in". Many exogenous (extraterrestrial) forces are caused by other bodies in the Solar System.

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Why is energy required for the heating or boiling process?

guapka [62]

To overcome the forces of attraction
intermolecular forces of attraction for the liquids and gases

5 0

4 years ago

A balloon with a pressure of 2.2 atm and a volume of 2.5 L is released in the mountains of Colorado. As the balloon ascends, the

CaHeK987 [17]

Answer:

1.38 atm

Explanation:

Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant

P1V1= P2V2

P1 = 2.2 atm, V1 = 2.5L ,

P2 = ? , V2 = 4L

2.2 × 2.5 = P2 × 4

Divide both sides by 4

P2 = 5.5 ÷ 4

P2 = 1.38 atm

I hope this was helpful, please mark as brainliest

5 0

4 years ago

An unknown material has a mass of 5.75 g and a volume of 7.5 cm3.

Shtirlitz [24]

Hey there!:

Mass = 5.75 g

Volume = 7.5 cm³

Therefore:

Density = mass / volume

D = 5.75 / 7.5

D = 0.7 g/cm³

Answer C

Hope that helps!

3 0

3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M